Qstnid -Iv-8-a Conducting Loop Of Area 5.0 Cm2 Is Placed In A Magnetic Field Which Varies Sinusoidally With Time As B = B0 Sin Ωt Where B0 = 0.20 T And Ω = 300 S-1. The Normal To The Coil Makes An Angle Of 60° With The Field. Find (A) The Maximum Emf Induced In The Coil, (B) The Emf Induced At Τ = (Π/900)s And (C) The Emf Induced At T = (Π/600) S. (A) The Maximum Emf Induced In The Coil, (B) The Emf Induced At Τ = (Π/900)s And (C) The Emf Induced At T = (Π/600) S. - 38: Electromagnetic Induction

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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 4

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#8
A conducting loop of area 5.0 cm² is placed in a magnetic field which varies sinusoidally with time as B = B₀ sin ωt where B₀ = 0.20 T and ω = 300 s^-1. The normal to the coil makes an angle of 60° with the field. Find (a) the maximum emf induced in the coil, (b) the emf induced at τ = (π/900)s and (c) the emf induced at t = (π/600) s. (a) the maximum emf induced in the coil, (b) the emf induced at τ = (π/900)s and (c) the emf induced at t = (π/600) s.
Ans : Given:
Area of the coil, A = 5 cm² = 5 × 10^-4 m²
The magnetic field at time t is given by
B = B₀ sin ωt = 0.2 sin (300t)
Angle of the normal of the coil with the magnetic field, θ = 60° (a) The emf induced in the coil is given by
`` e=\frac{-d\theta }{dt}=\frac{d}{dt}(BA\,\mathrm{\,cos\,}\theta )``
`` =\frac{d}{dt}\left[\left({B}_{0}\,\mathrm{\,sin\,}\,\mathrm{\,\omega \,}t\right)\times 5\times {10}^{-4}\times 1/2\right]``
`` ={B}_{0}\times \frac{5}{2}\times {10}^{-4}\frac{d}{dt}(\,\mathrm{\,sin\,}\,\mathrm{\,\omega \,}t)``
`` =\frac{{B}_{0}5}{2}{10}^{-4}\omega \left(\,\mathrm{\,cos\,}\,\mathrm{\,\omega \,}t\right)``
`` =\frac{0.2\times 5}{2}\times 300\times {10}^{-4}\times \,\mathrm{\,cos\,}\,\mathrm{\,\omega \,}t``
`` =15\times {10}^{-3}\,\mathrm{\,cost\,}\,\mathrm{\,\omega \,}t``
The induced emf becomes maximum when cos ωt becomes maximum, that is, 1.
Thus, the maximum value of the induced emf is given by
`` {e}_{max}=15\times {10}^{-3}=0.015\,\mathrm{\,V\,}`` (b) The induced emf at t = `` \left(\frac{\,\mathrm{\,\pi \,}}{900}\right)\,\mathrm{\,s\,}`` is given by
e = 15 × 10^-3 × cos ωt
= 15 × 10^-3 × cos `` \left(300\times \frac{\,\mathrm{\,\pi \,}}{900}\right)``
= 15 × 10^-3 × `` \frac{1}{2}``
`` =\frac{0.015}{2}=0.0075=7.5\times {10}^{-3}\,\mathrm{\,V\,}`` (c) The induced emf at t = `` \frac{\,\mathrm{\,\pi \,}}{600}\,\mathrm{\,s\,}`` is given by
e = 15 × 10^-3 × cos `` \left(300\times \frac{\,\mathrm{\,\pi \,}}{600}\right)``
= 15 × 10^-3× 0 = 0 V
Page No 306: (a) The emf induced in the coil is given by
`` e=\frac{-d\theta }{dt}=\frac{d}{dt}(BA\,\mathrm{\,cos\,}\theta )``
`` =\frac{d}{dt}\left[\left({B}_{0}\,\mathrm{\,sin\,}\,\mathrm{\,\omega \,}t\right)\times 5\times {10}^{-4}\times 1/2\right]``
`` ={B}_{0}\times \frac{5}{2}\times {10}^{-4}\frac{d}{dt}(\,\mathrm{\,sin\,}\,\mathrm{\,\omega \,}t)``
`` =\frac{{B}_{0}5}{2}{10}^{-4}\omega \left(\,\mathrm{\,cos\,}\,\mathrm{\,\omega \,}t\right)``
`` =\frac{0.2\times 5}{2}\times 300\times {10}^{-4}\times \,\mathrm{\,cos\,}\,\mathrm{\,\omega \,}t``
`` =15\times {10}^{-3}\,\mathrm{\,cost\,}\,\mathrm{\,\omega \,}t``
The induced emf becomes maximum when cos ωt becomes maximum, that is, 1.
Thus, the maximum value of the induced emf is given by
`` {e}_{max}=15\times {10}^{-3}=0.015\,\mathrm{\,V\,}`` (b) The induced emf at t = `` \left(\frac{\,\mathrm{\,\pi \,}}{900}\right)\,\mathrm{\,s\,}`` is given by
e = 15 × 10^-3 × cos ωt
= 15 × 10^-3 × cos `` \left(300\times \frac{\,\mathrm{\,\pi \,}}{900}\right)``
= 15 × 10^-3 × `` \frac{1}{2}``
`` =\frac{0.015}{2}=0.0075=7.5\times {10}^{-3}\,\mathrm{\,V\,}`` (c) The induced emf at t = `` \frac{\,\mathrm{\,\pi \,}}{600}\,\mathrm{\,s\,}`` is given by
e = 15 × 10^-3 × cos `` \left(300\times \frac{\,\mathrm{\,\pi \,}}{600}\right)``
= 15 × 10^-3× 0 = 0 V
Page No 306:

#8-a
the maximum emf induced in the coil,
Ans : The emf induced in the coil is given by
`` e=\frac{-d\theta }{dt}=\frac{d}{dt}(BA\,\mathrm{\,cos\,}\theta )``
`` =\frac{d}{dt}\left[\left({B}_{0}\,\mathrm{\,sin\,}\,\mathrm{\,\omega \,}t\right)\times 5\times {10}^{-4}\times 1/2\right]``
`` ={B}_{0}\times \frac{5}{2}\times {10}^{-4}\frac{d}{dt}(\,\mathrm{\,sin\,}\,\mathrm{\,\omega \,}t)``
`` =\frac{{B}_{0}5}{2}{10}^{-4}\omega \left(\,\mathrm{\,cos\,}\,\mathrm{\,\omega \,}t\right)``
`` =\frac{0.2\times 5}{2}\times 300\times {10}^{-4}\times \,\mathrm{\,cos\,}\,\mathrm{\,\omega \,}t``
`` =15\times {10}^{-3}\,\mathrm{\,cost\,}\,\mathrm{\,\omega \,}t``
The induced emf becomes maximum when cos ωt becomes maximum, that is, 1.
Thus, the maximum value of the induced emf is given by
`` {e}_{max}=15\times {10}^{-3}=0.015\,\mathrm{\,V\,}``

#8-b
the emf induced at τ = (π/900)s and
Ans : The induced emf at t = `` \left(\frac{\,\mathrm{\,\pi \,}}{900}\right)\,\mathrm{\,s\,}`` is given by
e = 15 × 10^-3 × cos ωt
= 15 × 10^-3 × cos `` \left(300\times \frac{\,\mathrm{\,\pi \,}}{900}\right)``
= 15 × 10^-3 × `` \frac{1}{2}``
`` =\frac{0.015}{2}=0.0075=7.5\times {10}^{-3}\,\mathrm{\,V\,}``

#8-c
the emf induced at t = (π/600) s.
Ans : The induced emf at t = `` \frac{\,\mathrm{\,\pi \,}}{600}\,\mathrm{\,s\,}`` is given by
e = 15 × 10^-3 × cos `` \left(300\times \frac{\,\mathrm{\,\pi \,}}{600}\right)``
= 15 × 10^-3× 0 = 0 V
Page No 306: